# Tag Info

Hint: $\pi$ is not a Liouville number, so there exists $m\in\mathbb{N}$ such that for all $p,q\in\mathbb{Z}$ with $q>1$, we have $$\left| \pi - \frac{p}{q}\right| \geq \frac{1}{q^m}.$$ This should allow you to keep $\sin n$ away from 0. Edit: Full Solution: Let $m$ be as above. So for all $p,q\in\mathbb{Z}$ with $q>1$ we have $$\left| \pi - ... 16 Supose not, so that there exists an \varepsilon>0 such that (0,\varepsilon)\cap S=\emptyset. \qquad\qquad\qquad(\star) It follows that \alpha=\inf S\cap(0,+\infty) is a positive number. The choice of \alpha and its positivity implies that the one and only element of S which is in [0,\alpha) is 0. I claim that \alpha\in S. ... 15 This isn't exceptionally good compared to the partial convergents of the continued fraction expansion. Terminating the continued fraction for \pi right before the 292 gives \frac{355}{113}= \textbf{3.141592}9203\ldots, which gets 6 digits after the decimal place right, while your fraction only gets 3 digits after the decimal place correct even ... 10 Though it is clear what would count as a positive answer, the question is not precise enough that to make clear what counts as a negative answer. Nonetheless, I believe that no similar formula could possibly be equivalent to denseness. The problem is that there are sequences that are dense but that are wildly non-equidistributed. For example, your formula ... 10 Using the property stated in that article:$$\mu(x)=2 + \limsup \frac{\log a_{n+1}}{\log q_n}$$where the continued fraction expansion for x is [a_0,a_1,...] and the nth convergent is \frac{p_n}{q_n}. Start with a_0=2 and a_1=2, so q_0=1, q_1=2. Now, assume you have a continued fraction$$\frac{p_n}{q_n}=[a_0,...,a_n]$$Define a_{n+1} ... 9 Actually, we can do it for higher powers as well. Let F_k (N) denote the sum of the k^{th} powers of the differences in the Farey sequence. Then it is easy to see that$$F_0(N)=\sum_{n=1}^N \phi(n)\sim \frac{3N^2}{\pi^2}$$and$$F_1(N)=1.$$It seems you are looking for F_2 (N), and, you are correct that the asymptotic is \frac{\log N}{N^2}. More ... 8 Suppose that the prime factors of n^2+1 are all bounded by N for infinitely many n. Then infinitely many integers n^2 + 1 can be written in the form D y^3 for one of finitely many D. Explicitly, the set of D can be taken to be the finitely many integers whose prime divisors are all less than N, and whose exponents are at most 2. For ... 8 There are some details that I haven't fully vetted but here's a long sketch which I believe should show divergence. Define c_n := \frac{n}{2\pi} \pmod 1 and let [a,b] be any interval in the torus \mathbb R/\mathbb Z. The discrepancy D(N) of the sequence c_n is defined to be the difference between \# \{ n \le N : c_n \in [a,b]\} and the expected ... 7 I proceed the same way as in my other answer here http://math.stackexchange.com/a/110019/17445 |\sin n| \le \varepsilon implies that there exists an integer k(n) such that n = k(n)\pi + a(n) where a(n) \in [-\arcsin(\varepsilon) ; \arcsin(\varepsilon)] \subset [-\pi \varepsilon/2 ; \pi \varepsilon/2], and if both |\sin n| and |\sin m| are less ... 7 Let x be a fixed positive real. Then, the behavior of the sequence \{ kx \} falls under one of the following three types, depending on x. If x is an integer: In this case, the sequence \{ kx \} is the constant sequence 0. In particular, the sequence converges to 0. If x is rational, but not an integer: Suppose x = a/b where a,b are ... 7 The problem is that equidistribution is a property of sequences, but density is a property of sets. You may wish to prove as an exercise that a countable subset of [0,1] is dense if and only if it can be ordered in such a way as to be equidistributed. So the set \lbrace\,x_0,x_1,\dots\,\rbrace is dense if and only if there is a permutation \sigma such ... 7 Assume that a, b \in \mathbb N  are not perfect squares and  \sqrt a + \sqrt b = n \in \mathbb N. Then$$ a = (n - \sqrt b)^2 = n^2 - 2 n \sqrt b + b $$which means that \sqrt b is a rational number. This contradicts the fact that the square root of an integer is either an integer or a irrational number. 7 Suppose a+b=c, so that a+b-c=0, with a^3, b^3, c all rational. Then we have -3abc=a^3+b^3-c^3 by virtue of the identity$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$(take c with a negative sign) Hence a+b and ab are both rational, so a and b satisfy a quadratic equation with rational coefficients. There are lots of ways of ... 7 normal \; \not\Rightarrow \; irrationality measure 2 There exist numbers that are normal with irrationality measure >2. In fact, there exist normal numbers (meaning normal with respect to every base) that have irrationality measure \infty. This is Theorem 2 in Bugeaud [2] (2002). For related results, see [1] and [3]. irrationality measure 2 ... 6 You can determine this by looking at the continued fractions of 0.8575 and 0.8585. They are [0, 1, 6, 57] and [0, 1, 6, 14, 1, 8, 2] so anything in [0, 1, 6, 15..57] would work. [0, 1, 6, 15] has the smallest denominator in that range which corresponds to 0+1/(1+1/(6+1/15)) = 91/106. I just tested smaller denominators and this checks out right. ... 5 The Weyl equidistribution theorem says that for irrational \alpha and sufficiently many k, the fractional parts \{\alpha k\} will be equidistributed in the interval [0,1]. To apply this to your particular problem, consider a large interval k\in[N,2N]. Because of equidistribution, the numbers k will "hit" the condition \{\alpha k\} < ... 5 Taking a step into generalization, it is true that every additive subgroup G of \mathbb R is either discrete or dense. This can be proved by considering \alpha = \inf \{ x \in G : x>0 \}. Then G is discrete iff \alpha >0, in which case G=\alpha \mathbb Z. In your case, Jyrki's suggestion implies that \alpha=0 and so S is dense. 5 If we look at the continued fraction of the expression \frac{\pi}{180}, we get:$$\frac\pi{180} = [0,57,3,2,1,1,1,2,40,\dots]$$And:$$\frac{44}{2521} = [0,57,3,2,1,1,1,1]$$So that's a pretty good match. It is between two terms of the continued fraction expansion of \frac\pi{180}, in particular. Now, whether this is surprisingly good in any way, ... 4 A good way to construct fractions which approximate numbers is to use continued fractions. Namely, if we have that x=a_0+1/(a_1+1/(a_2+1/(\cdots (which we shall write with the shorthand x=[a_0,a_1,a_2,\ldots]), we can define the convergents of x to be the partial sums [a_1, \ldots, a_k]=a_0+1/(a_1 + 1/(\ldots +1/(a_k))\cdots)). These will give the ... 4 Some 'brute force' solutions of \ x^3 + x + y^3 + y - z^3 - z = r (with additional contributions from Oleg567) For \ r=0\  (and z<10^5) : \begin{array} {cc|c} x&y&z\\ \hline 36& 37& 46\\ 98& 248& 253\\ 165& 705& 708\\ 320& 377& 442\\ 843& 1078& 1228\\ 2372& 3323& 3685\\ 2988& 3070& ... 3 The proof for solutions of \displaystyle{2x^2+1=3^n} can be read from the paper at American Mathematical Society Volume 131, Number 12 According to that three solutions are (1,1), (2,2) and (11,5) NOTE: I believe one cannot attempt with one approach to solve all of those equations. ADDING THESE NOTE (Since it was requested in the comment here) A ... 3 If you know how to solve problems of the form "find the least integer n(\eta) \ge 1 such that n\alpha \in [- \eta - \varepsilon; - \eta + \varepsilon] \pmod 1" where |\eta| \le \varepsilon, then you can easily find all those integers one after the other. Thus, you want to keep a list of the integers n such that n \alpha is closer to 0 than for ... 3 The statement can be false even when you assume \frac{\theta}{\pi} is irrational. Let (s_k)_{k=1,2\ldots} be the sequence defined recursively by:$$s_k = \begin{cases}1,&\quad\text{ for }k = 1\\2^{s_{k-1}},&\quad\text{ for }k > 1\end{cases}$$and s be the Liouville number s = \sum_{k=1}^{\infty} 2^{-s_k}. For r = \sqrt{2}, and \theta ... 3 A proof of the fact that the set of reals with irrationality measure \gt 2 has Lebesgue measure 0 can be found in several places. I think the result is due to Khinchin, and is in his book on continued fractions. Here is an online proof by Beukers. It is towards the end of the paper, but independent of the rest, which anyway is worth reading. It ... 3 This is a partial sum of a geometric series:$$\left| \frac{1}{n} \sum_{m=1}^n e^{2 \pi i h (2 \pi m)} \right| =\left| \frac{1}{n} \sum_{m=1}^n q^m \right| =\left| \frac{q}{n} \frac{1-q^n}{1-q}\right| \le \frac{2}{n\left|1-q\right|}$$with q:=e^{(2 \pi)^2 i h}. If your question is whether there is an upper bound for this independent of h, the ... 3 Here are some necessary and sufficient conditions: 1.a. The set \{n\mid a\leqslant x_n\leqslant b\} is nonempty for every 0\leqslant a<b\leqslant 1. 1.b. The set \{n\mid a\leqslant x_n\leqslant b\} is infinite for every 0\leqslant a<b\leqslant 1. 2.a. The sum of the series \sum\limits_nf(x_n) is positive for every continuous nonnegative ... 3 If \left|\frac pq-r\right|\lt\frac1{2q^2}, then \frac pq is a convergent for the continued fraction for r\not\in\mathbb{Q}. If \frac pq is a convergent for the continued fraction for r\not\in\mathbb{Q}, then \left|\frac pq-r\right|\lt\frac1{q^2} Thus, I would call \frac pq is a "good" approximation for r if \left|\frac ... 3$$|\sqrt{2} - \frac{m}{n}| < \frac{1}{4n^2}\\\implies |\sqrt{2}n-m|< \frac{1}{4n}\\\implies \sqrt{2}n-\frac{1}{4n}<m<\sqrt{2}n+\frac{1}{4n}\\\implies 0<|(\sqrt{2}n-m)(\sqrt{2}n+m)|<\frac{1}{4n}(2\sqrt{2}n+\frac{1}{4n})=\frac{\sqrt{2}}{2}+\frac{1}{16n^2}<1. But $|(\sqrt{2}n-m)(\sqrt{2}n+m)|=|2n^2-m^2|$ is an integer, a contradiction.
The flaw is that you define $h$ as $m+\varepsilon/n$, which is not necessarily an integer. Also, the theorem does not imply that the rational numbers can not be arbitrarily approximated by rational numbers, but that they can not be arbitrarily approximated with "small" denominators. This is because - as you already noted - the error in the approximation is ...